Chapter 1
Units associated with basic
electrical quantities
At the end of this chapter you should be able to:
• state the basic SI units
• recognize derived SI units
• understand prefixes denoting multiplication and division
• state the units of charge, force, work and power and perform simple calculations involving these units
• state the units of electrical potential, e.m.f., resistance, conductance, power and energy and perform simple
calculations involving these units
1.1
Derived SI units use combinations of basic units and
there are many of them. Two examples are:
SI units
The system of units used in engineering and science is
the Système Internationale d’Unités (International system of units), usually abbreviated to SI units, and is
based on the metric system. This was introduced in 1960
and is now adopted by the majority of countries as the
official system of measurement.
The basic units in the SI system are listed below with
their symbols:
Velocity – metres per second (m/s)
Acceleration – metres per second
squared (m/s2 )
SI units may be made larger or smaller by using prefixes
which denote multiplication or division by a particular
amount. The six most common multiples, with their
meaning, are listed below:
Prefix
Name
Meaning
M
mega
multiply by 1 000 000 (i.e.×106 )
k
kilo
multiply by 1000 (i.e. ×103 )
second, s
m
milli
divide by 1000 (i.e. ×10−3 )
electric current
ampere, A
μ
micro
divide by 1 000 000 (i.e. ×10−6 )
thermodynamic temperature
kelvin, K
n
nano
divide by 1 000 000 000
(i.e. ×10−9 )
luminous intensity
candela, cd
p
pico
amount of substance
mole, mol
divide by 1 000 000 000 000
(i.e. ×10−12 )
Quantity
Unit
length
metre, m
mass
kilogram, kg
time
Section 1
4 Electrical and Electronic Principles and Technology
1.2 Charge
1.4 Work
The unit of charge is the coulomb (C) where one
coulomb is one ampere second. (1 coulomb = 6.24 ×
1018 electrons). The coulomb is defined as the quantity
of electricity which flows past a given point in an electric circuit when a current of one ampere is maintained
for one second. Thus,
charge, in coulombs Q = It
where I is the current in amperes and t is the time in
seconds.
Problem 1. If a current of 5 A flows for 2 minutes,
find the quantity of electricity transferred.
Quantity of electricity Q = It coulombs
Hence
1.3
I = 5 A, t = 2 × 60 = 120 s
Q = 5 × 120 = 600 C
The unit of work or energy is the joule (J) where one
joule is one newton metre. The joule is defined as the
work done or energy transferred when a force of one
newton is exerted through a distance of one metre in the
direction of the force. Thus
work done on a body, in joules, W = Fs
where F is the force in newtons and s is the distance in
metres moved by the body in the direction of the force.
Energy is the capacity for doing work.
1.5
Power
The unit of power is the watt (W) where one watt is one
joule per second. Power is defined as the rate of doing
work or transferring energy. Thus,
W
t
where W is the work done or energy transferred, in
joules, and t is the time, in seconds. Thus,
power, in watts,
Force
The unit of force is the newton (N) where one newton
is one kilogram metre per second squared. The newton
is defined as the force which, when applied to a mass of
one kilogram, gives it an acceleration of one metre per
second squared. Thus,
force, in newtons
F = ma
where m is the mass in kilograms and a is the acceleration in metres per second squared. Gravitational force,
or weight, is mg, where g = 9.81 m/s2 .
Problem 2. A mass of 5000 g is accelerated at
2 m/s2 by a force. Determine the force needed.
Force = mass × acceleration
= 5 kg × 2 m/s2 = 10 kg m/s2 = 10 N.
Problem 3. Find the force acting vertically
downwards on a mass of 200 g attached to a wire.
Mass = 200 g = 0.2 kg and acceleration due to gravity,
g = 9.81 m/s2
Force acting
= weight
downwards
= mass × acceleration
= 0.2 kg × 9.81 m/s2
= 1.962 N
P=
energy, in joules, W = Pt
Problem 4. A portable machine requires a force
of 200 N to move it. How much work is done if the
machine is moved 20 m and what average power is
utilized if the movement takes 25 s?
Work done = force × distance
= 200 N × 20 m
= 4 000 Nm or 4 kJ
work done
Power =
time taken
=
4000 J
= 160 J/s = 160 W
25 s
Problem 5. A mass of 1000 kg is raised through a
height of 10 m in 20 s. What is (a) the work done
and (b) the power developed?
(a) Work done = force × distance
and force = mass × acceleration
Hence,
distance of 1.5 cm in 40 ms. Find the power
consumed.
[4.5 W]
work done = (1000 kg × 9.81 m/s2 ) × (10 m)
= 98 100 Nm
= 98.1 kNm or 98.1 kJ
(b)
work done
98100 J
=
time taken
20 s
= 4905 J/s = 4905 W or 4.905 kW
Power =
Now try the following exercise
Exercise 1 Further problems on charge,
force, work and power
(Take g = 9.81 m/s2 where appropriate)
12. A mass of 500 kg is raised to a height of 6 m
in 30 s. Find (a) the work done and (b) the
power developed.
[(a) 29.43 kNm (b) 981 W]
13. Rewrite the following as indicated:
(a) 1000 pF = . . . . . . nF
(b) 0.02 μF = . . . . . . pF
(c) 5000 kHz = . . . . . . MHz
(d) 47 k = . . . . . . M
(e) 0.32 mA = . . . . . . μA
[(a) 1 nF (b) 20000 pF (c) 5 MHz
(d) 0.047 M (e) 320 μA]
1. What quantity of electricity is carried by
6.24 × 1021 electrons?
[1000 C]
2. In what time would a current of 1 A transfer
a charge of 30 C?
[30 s]
3. A current of 3 A flows for 5 minutes. What
charge is transferred?
[900 C]
4. How long must a current of 0.1 A flow so as
to transfer a charge of 30 C?
[5 minutes]
1.6
The unit of electric potential is the volt (V), where one
volt is one joule per coulomb. One volt is defined as
the difference in potential between two points in a conductor which, when carrying a current of one ampere,
dissipates a power of one watt, i.e.
watts
joules/second
=
amperes
amperes
joules
joules
=
=
ampere seconds
coulombs
5. What force is required to give a mass of 20 kg
an acceleration of 30 m/s2 ?
[600 N]
6. Find the accelerating force when a car having
a mass of 1.7 Mg increases its speed with a
constant acceleration of 3 m/s2 .
[5.1 kN]
5 m/s2 .
7. A force of 40 N accelerates a mass at
Determine the mass.
[8 kg]
8. Determine the force acting downwards on
a mass of 1500 g suspended on a string.
[14.72 N]
9. A force of 4 N moves an object 200 cm in the
direction of the force. What amount of work
is done?
[8 J]
10. A force of 2.5 kN is required to lift a load.
How much work is done if the load is lifted
through 500 cm?
[12.5 kJ]
11. An electromagnet exerts a force of 12 N
and moves a soft iron armature through a
Electrical potential and e.m.f.
volts =
A change in electric potential between two points in
an electric circuit is called a potential difference. The
electromotive force (e.m.f.) provided by a source of
energy such as a battery or a generator is measured in
volts.
1.7
Resistance and conductance
The unit of electric resistance is the ohm(), where
one ohm is one volt per ampere. It is defined as the
resistance between two points in a conductor when a
constant electric potential of one volt applied at the
two points produces a current flow of one ampere in
the conductor. Thus,
resistance, in ohms
R=
V
I
Section 1
Units associated with basic electrical quantities 5
Section 1
6 Electrical and Electronic Principles and Technology
where V is the potential difference across the two points,
in volts, and I is the current flowing between the two
points, in amperes.
The reciprocal of resistance is called conductance
and is measured in siemens (S). Thus
conductance, in siemens G =
1
R
where R is the resistance in ohms.
Problem 6. Find the conductance of a conductor
of resistance: (a) 10 (b) 5 k (c) 100 m.
(a) Conductance G =
1
1
=
siemen = 0.1 S
R 10
(b) G =
1
1
=
S = 0.2 × 10−3 S = 0.2 mS
R 5 × 103
(c) G =
1
103
1
S=
=
S = 10 S
−3
R 100 × 10
100
1.8
Electrical power and energy
When a direct current of I amperes is flowing in an
electric circuit and the voltage across the circuit is
V volts, then
power, in watts P = VI
Electrical energy = Power × time
= VIt joules
Although the unit of energy is the joule, when dealing with large amounts of energy, the unit used is the
kilowatt hour (kWh) where
1 kWh = 1000 watt hour
= 1000 × 3600 watt seconds or joules
= 3 600 000 J
Problem 7. A source e.m.f. of 5 V supplies a
current of 3 A for 10 minutes. How much energy is
provided in this time?
Energy = power × time, and power = voltage × current.
Hence
Energy = VIt = 5 × 3 × (10 × 60)
= 9000 Ws or J = 9 kJ
Problem 8. An electric heater consumes 1.8 MJ
when connected to a 250 V supply for 30 minutes.
Find the power rating of the heater and the current
taken from the supply.
Power =
energy 1.8 × 106 J
=
time
30 × 60 s
= 1000 J/s = 1000 W
i.e. power rating of heater = 1 kW
Power P = VI, thus I =
P 1000
=
=4A
V
250
Hence the current taken from the supply is 4 A.
Now try the following exercise
Exercise 2 Further problems on e.m.f., resistance, conductance, power and
energy
1. Find the conductance of a resistor of
resistance (a) 10 (b) 2 k (c) 2 m
[(a) 0.1 S (b) 0.5 mS (c) 500 S]
2. A conductor has a conductance of 50 μS. What
is its resistance?
[20 k]
3. An e.m.f. of 250 V is connected across a
resistance and the current flowing through
the resistance is 4 A. What is the power
developed?
[1 kW]
4. 450 J of energy are converted into heat in
1 minute. What power is dissipated?
[7.5 W]
5. A current of 10 A flows through a conductor
and 10 W is dissipated. What p.d. exists across
the ends of the conductor?
[1 V]
6. A battery of e.m.f. 12 V supplies a current
of 5 A for 2 minutes. How much energy is
supplied in this time?
[7.2 kJ]
7. A d.c. electric motor consumes 36 MJ when
connected to a 250 V supply for 1 hour. Find
the power rating of the motor and the current
taken from the supply.
[10 kW, 40 A]
1.9
Summary of terms, units and
their symbols
2. Complete the following:
Force = . . . . . . × . . . . . .
Quantity
Quantity Unit
Symbol
Unit
Symbol
Length
l
metre
m
Mass
m
kilogram
kg
Time
t
second
s
Velocity
v
metres per m/s or
second
m s−1
Acceleration
a
metres per m/s2 or
second
m s−2
squared
Force
F
newton
N
Electrical
charge or
quantity
Q
coulomb
C
3. What do you understand by the term ‘potential difference’?
4. Define electric current in terms of charge and
time
5. Name the units used to measure:
(a) the quantity of electricity
(b) resistance
(c) conductance
6. Define the coulomb
7. Define electrical energy and state its unit
8. Define electrical power and state its unit
9. What is electromotive force?
Electric current I
ampere
A
Resistance
R
ohm
Conductance
G
siemen
S
Electromotive
force
E
volt
V
Potential
difference
V
volt
V
Work
W
joule
J
Energy
E (or W)
joule
J
Power
P
watt
W
Now try the following exercises
Exercise 3 Short answer questions on units
associated with basic electrical
quantities
1. What does ‘SI units’ mean?
10. Write down a formula for calculating the
power in a d.c. circuit
11. Write down the symbols for the following
quantities:
(a) electric charge
(b) work
(c) e.m.f.
(d) p.d.
12. State which units the following abbreviations
refer to:
(a) A
(b) C
(c) J
(d) N
(e) m
Exercise 4 Multi-choice questions on units
associated with basic electrical
quantities (Answers
on page 398)
1. A resistance of 50 k has a conductance of:
(a) 20 S
(b) 0.02 S
(c) 0.02 mS
(d) 20 kS
2. Which of the following statements is
incorrect?
(a) 1 N = 1 kg m/s2
(b) 1 V = 1 J/C
(c) 30 mA = 0.03 A
(d) 1 J = 1 N/m
Section 1
Units associated with basic electrical quantities 7
Section 1
8 Electrical and Electronic Principles and Technology
3. The power dissipated by a resistor of 10 when a current of 2 A passes through it is:
(a) 0.4 W (b) 20 W (c) 40 W (d) 200 W
4. A mass of 1200 g is accelerated at 200 cm/s2
by a force. The value of the force required is:
(a) 2.4 N
(b) 2,400 N
(c) 240 kN
(d) 0.24 N
5. A charge of 240 C is transferred in 2 minutes.
The current flowing is:
(a) 120 A (b) 480 A (c) 2 A (d) 8 A
6. A current of 2 A flows for 10 h through a
100 resistor. The energy consumed by the
resistor is:
(a) 0.5 kWh
(b) 4 kWh
(c) 2 kWh
(d) 0.02 kWh
7. The unit of quantity of electricity is the:
(a) volt
(b) coulomb
(c) ohm
(d) joule
8. Electromotive force is provided by:
(a) resistance’s
(b) a conducting path
(c) an electric current
(d) an electrical supply source
9. The coulomb is a unit of:
(a) power
(b) voltage
(c) energy
(d) quantity of electricity
10. In order that work may be done:
(a) a supply of energy is required
(b) the circuit must have a switch
(c) coal must be burnt
(d) two wires are necessary
11. The ohm is the unit of:
(a) charge
(c) power
(b) resistance
(d) current
12. The unit of current is the:
(a) volt
(b) coulomb
(c) joule
(d) ampere
Chapter 2
An introduction to
electric circuits
At the end of this chapter you should be able to:
• appreciate that engineering systems may be represented by block diagrams
• recognize common electrical circuit diagram symbols
• understand that electric current is the rate of movement of charge and is measured in amperes
• appreciate that the unit of charge is the coulomb
• calculate charge or quantity of electricity Q from Q = It
•
•
•
•
•
•
understand that a potential difference between two points in a circuit is required for current to flow
appreciate that the unit of p.d. is the volt
•
•
•
•
•
calculate electrical power
define electrical energy and state its unit
calculate electrical energy
state the three main effects of an electric current, giving practical examples of each
understand that resistance opposes current flow and is measured in ohms
appreciate what an ammeter, a voltmeter, an ohmmeter, a multimeter and an oscilloscope measure
distinguish between linear and non-linear devices
state Ohm’s law as V = IR or I = V /R or R = V /I
• use Ohm’s law in calculations, including multiples and sub-multiples of units
• describe a conductor and an insulator, giving examples of each
• appreciate that electrical power P is given by P = VI = I 2 R = V 2 /R watts
explain the importance of fuses in electrical circuits
2.1
Electrical/electronic system
block diagrams
An electrical/electronic system is a group of components connected together to perform a desired function.
Figure 2.1 shows a simple public address system, where
a microphone is used to collect acoustic energy in
the form of sound pressure waves and converts this
to electrical energy in the form of small voltages
and currents; the signal from the microphone is then
amplified by means of an electronic circuit containing
transistors/integrated circuits before it is applied to the
loudspeaker.
10 Electrical and Electronic Principles and Technology
Section 1
A.C. Supply
Microphone
Loudspeaker
Amplifier
Enclosure
Temperature
of enclosure
Figure 2.3
Figure 2.1
A sub-system is a part of a system which performs
an identified function within the whole system; the
amplifier in Fig. 2.1 is an example of a sub-system.
A component or element is usually the simplest
part of a system which has a specific and well-defined
function – for example, the microphone in Fig. 2.1.
The illustration in Fig. 2.1 is called a block diagram
and electrical/electronic systems, which can often be
quite complicated, can be better understood when broken down in this way. It is not always necessary to know
precisely what is inside each sub-system in order to
know how the whole system functions.
As another example of an engineering system,
Fig. 2.2 illustrates a temperature control system containing a heat source (such as a gas boiler), a fuel controller
(such as an electrical solenoid valve), a thermostat and
a source of electrical energy. The system of Fig. 2.2 can
be shown in block diagram form as in Fig. 2.3; the thermostat compares the actual room temperature with the
desired temperature and switches the heating on or off.
240 V
Gas
boiler
Solenoid
Thermostat
Error
Heating
+
system
Temperature
−
command
Actual
temperature
network could comprise a file server, coaxial cable, network adapters, several computers and a laser printer; an
electromechanical system is another example, where a
car electrical system could comprise a battery, a starter
motor, an ignition coil, a contact breaker and a distributor. All such systems as these may be represented by
block diagrams.
2.2
Standard symbols for electrical
components
Symbols are used for components in electrical circuit
diagrams and some of the more common ones are shown
in Fig. 2.4.
Conductor
Two conductors
crossing but not
joined
Two conductors
joined together
Fixed resistor
Alternative symbol
for fixed resistor
Variable resistor
Cell
Battery of 3 cells
Alternative symbol
for battery
Switch
Filament lamp
Fuse
A
V
Ammeter
Voltmeter
Fuel
supply
Thermostat
Set temperature
Radiators
Enclosed space
Figure 2.2
There are many types of engineering systems. A communications system is an example, where a local area
Figure 2.4
Alternative fuse
symbol
2.3
Electric current and quantity of
electricity
All atoms consist of protons, neutrons and electrons.
The protons, which have positive electrical charges, and
the neutrons, which have no electrical charge, are contained within the nucleus. Removed from the nucleus
are minute negatively charged particles called electrons.
Atoms of different materials differ from one another by
having different numbers of protons, neutrons and electrons. An equal number of protons and electrons exist
within an atom and it is said to be electrically balanced,
as the positive and negative charges cancel each other
out. When there are more than two electrons in an atom
the electrons are arranged into shells at various distances
from the nucleus.
All atoms are bound together by powerful forces of
attraction existing between the nucleus and its electrons. Electrons in the outer shell of an atom, however,
are attracted to their nucleus less powerfully than are
electrons whose shells are nearer the nucleus.
It is possible for an atom to lose an electron; the atom,
which is now called an ion, is not now electrically balanced, but is positively charged and is thus able to attract
an electron to itself from another atom. Electrons that
move from one atom to another are called free electrons and such random motion can continue indefinitely.
However, if an electric pressure or voltage is applied
across any material there is a tendency for electrons to
move in a particular direction. This movement of free
electrons, known as drift, constitutes an electric current
flow. Thus current is the rate of movement of charge.
Conductors are materials that contain electrons that
are loosely connected to the nucleus and can easily move
through the material from one atom to another.
Insulators are materials whose electrons are held
firmly to their nucleus.
The unit used to measure the quantity of electrical charge Q is called the coulomb C (where
1 coulomb = 6.24 × 1018 electrons)
If the drift of electrons in a conductor takes place at
the rate of one coulomb per second the resulting current
is said to be a current of one ampere.
1 ampere = 1 coulomb per second or
1 A = 1 C/s
Hence 1 coulomb = 1 ampere second or
1 C = 1 As
Thus
Generally, if I is the current in amperes and t the time
in seconds during which the current flows, then I × t
represents the quantity of electrical charge in coulombs,
i.e. quantity of electrical charge transferred,
Q = I × t coulombs
Problem 1. What current must flow if 0.24
coulombs is to be transferred in 15 ms?
Since the quantity of electricity, Q = It, then
I=
Q
0.24
0.24 × 103
=
=
t
15 × 10−3
15
240
=
= 16 A
15
Problem 2. If a current of 10 A flows for four
minutes, find the quantity of electricity transferred.
Quantity of electricity, Q = It coulombs. I = 10 A and
t = 4 × 60 = 240 s. Hence
Q = 10 × 240 = 2400 C
Now try the following exercise
Exercise 5 Further problems on charge
1. In what time would a current of 10 A transfer
a charge of 50 C?
[5 s]
2. A current of 6 A flows for 10 minutes. What
charge is transferred?
[3600 C]
3. How long must a current of 100 mA flow so
as to transfer a charge of 80 C?
[13 min 20 s]
2.4 Potential difference and
resistance
For a continuous current to flow between two points in
a circuit a potential difference (p.d.) or voltage, V, is
required between them; a complete conducting path is
necessary to and from the source of electrical energy.
The unit of p.d. is the volt, V.
Figure 2.5 shows a cell connected across a filament
lamp. Current flow, by convention, is considered as
flowing from the positive terminal of the cell, around
the circuit to the negative terminal.
Section 1
An introduction to electric circuits 11
12 Electrical and Electronic Principles and Technology
Section 1
+
A
Current
flow
switch is on 10 V/cm then the magnitude of the p.d. is
3 cm × 10 V/cm, i.e. 30 V.
(See Chapter 10 for more detail about electrical
measuring instruments and measurements.)
2.6
V
Figure 2.5
The flow of electric current is subject to friction. This
friction, or opposition, is called resistance R and is the
property of a conductor that limits current. The unit of
resistance is the ohm; 1 ohm is defined as the resistance
which will have a current of 1 ampere flowing through
it when 1 volt is connected across it,
i.e.
resistance R =
Potential difference
current
Linear and non-linear devices
Figure 2.6 shows a circuit in which current I can be
varied by the variable resistor R2 . For various settings
of R2 , the current flowing in resistor R1 , displayed on
the ammeter, and the p.d. across R1 , displayed on the
voltmeter, are noted and a graph is plotted of p.d. against
current. The result is shown in Fig. 2.7(a) where the
straight line graph passing through the origin indicates
that current is directly proportional to the p.d. Since the
gradient, i.e. (p.d.)/(current) is constant, resistance R1
is constant. A resistor is thus an example of a linear
device.
V
2.5
A
Basic electrical measuring
instruments
An ammeter is an instrument used to measure current
and must be connected in series with the circuit. Figure 2.5 shows an ammeter connected in series with the
lamp to measure the current flowing through it. Since
all the current in the circuit passes through the ammeter
it must have a very low resistance.
A voltmeter is an instrument used to measure p.d. and
must be connected in parallel with the part of the circuit whose p.d. is required. In Fig. 2.5, a voltmeter is
connected in parallel with the lamp to measure the p.d.
across it. To avoid a significant current flowing through
it a voltmeter must have a very high resistance.
An ohmmeter is an instrument for measuring
resistance.
A multimeter, or universal instrument, may be used to
measure voltage, current and resistance. An ‘Avometer’
and ‘Fluke’ are typical examples.
The oscilloscope may be used to observe waveforms
and to measure voltages and currents. The display of
an oscilloscope involves a spot of light moving across
a screen. The amount by which the spot is deflected
from its initial position depends on the p.d. applied to
the terminals of the oscilloscope and the range selected.
The displacement is calibrated in ‘volts per cm’. For
example, if the spot is deflected 3 cm and the volts/cm
R1
l
R2
Figure 2.6
p.d.
p.d.
0
l
(a)
0
l
(b)
Figure 2.7
If the resistor R1 in Fig. 2.6 is replaced by a component such as a lamp then the graph shown in Fig. 2.7(b)
results when values of p.d. are noted for various current
readings. Since the gradient is changing, the lamp is an
example of a non-linear device.
or
2.7 Ohm’s law
From Ohm’s law, potential difference,
Ohm’s law states that the current I flowing in a circuit
is directly proportional to the applied voltage V and
inversely proportional to the resistance R, provided the
temperature remains constant. Thus,
I=
V
V
or V = IR or R =
R
I
10
10
A or
A = 0.01 A
3
1000
10
V = IR = (0.01)(2000) = 20 V
Problem 5. A coil has a current of 50 mA
flowing through it when the applied voltage is
12 V. What is the resistance of the coil?
Problem 3. The current flowing through a resistor
is 0.8 A when a p.d. of 20 V is applied. Determine
the value of the resistance.
Resistance, R =
From Ohm’s law,
=
V
20 200
resistance R = =
=
= 25 I
0.8
8
2.8
Multiples and sub-multiples
Currents, voltages and resistances can often be very
large or very small. Thus multiples and sub-multiples
of units are often used, as stated in Chapter 1. The most
common ones, with an example of each, are listed in
Table 2.1.
Problem 4. Determine the p.d. which must be
applied to a 2 k resistor in order that a current of
10 mA may flow.
12 × 103
12 000
=
= 240 50
50
Problem 6. A 100 V battery is connected across a
resistor and causes a current of 5 mA to flow.
Determine the resistance of the resistor. If the
voltage is now reduced to 25 V, what will be the
new value of the current flowing?
Resistance R =
V
100 × 103
100
=
=
−3
I
5 × 10
5
= 20 × 103 = 20 k
Current when voltage is reduced to 25 V,
Resistance R = 2 k = 2 × 103 = 2000 Current I = 10 mA = 10 × 10−3 A
Table 2.1
Prefix
V
12
=
I
50 × 10−3
I=
V
25
25
=
=
× 10−3 = 1.25 mA
R
20 × 103
20
Name
Meaning
Example
M
mega
multiply by 1 000 000
(i.e. ×106 )
2 M = 2 000 000 ohms
k
kilo
multiply by 1000
(i.e. ×103 )
10 kV = 10 000 volts
m
milli
divide by 1000
(i.e. ×10−3 )
25 mA =
μ
micro
divide by 1 000 000
(i.e. ×10−6 )
50 μV =
25
A
1000
= 0.025 amperes
50
V
1 000 000
= 0.000 05 volts
Section 1
An introduction to electric circuits 13
Problem 7. What is the resistance of a coil which
draws a current of (a) 50 mA and (b) 200 μA from a
120 V supply?
=
120 12 000
=
0.05
5
= 2400 or 2.4 k
(b) Resistance R =
=
Now try the following exercise
Exercise 6
Further problems on Ohm’s law
1. The current flowing through a heating element
is 5 A when a p.d. of 35 V is applied across it.
Find the resistance of the element.
[7 ]
V
120
(a) Resistance R = =
I
50 × 10−3
120
120
=
−6
200 × 10
0.0002
1 200 000
= 600 000 2
2. A 60 W electric light bulb is connected to a
240 V supply. Determine (a) the current flowing in the bulb and (b) the resistance of the
bulb.
[(a) 0.25 A (b) 960 ]
3. Graphs of current against voltage for two resistors P and Q are shown in Fig. 2.9 Determine
the value of each resistor.
[2 m, 5 m]
or 600 k or 0.6 M
25
P
10
8
Current/mA
Problem 8. The current/voltage relationship for
two resistors A and B is as shown in Fig. 2.8
Determine the value of the resistance of each
resistor.
6
Q
4
2
20
Current/mA
Section 1
14 Electrical and Electronic Principles and Technology
Resistor A
0
15
4
8
12
16
Voltage/μV
20
24
10
Figure 2.9
Resistor B
5
0
4
16
8
12
Voltage/V
20
4. Determine the p.d. which must be applied to a
5 k resistor such that a current of 6 mA may
flow.
[30 V]
5. A 20 V source of e.m.f. is connected across a
circuit having a resistance of 400 . Calculate
the current flowing.
[50 mA]
Figure 2.8
For resistor A,
R=
20 V
20
2000
V
=
=
=
I
20 mA
0.02
2
= 1000 or 1 k
For resistor B,
R=
16 V
16
16 000
V
=
=
=
I
5 mA
0.005
5
= 3200 or 3.2 k
2.9 Conductors and insulators
A conductor is a material having a low resistance which
allows electric current to flow in it. All metals are conductors and some examples include copper, aluminium,
brass, platinum, silver, gold and carbon.
An insulator is a material having a high resistance
which does not allow electric current to flow in it.
Some examples of insulators include plastic, rubber,
glass, porcelain, air, paper, cork, mica, ceramics and
certain oils.
2.10
Alternatively, since I = 4 × 10−3 and R = 5 × 103 then
from Ohm’s law, voltage
V = IR = 4 × 10−3 × 5 × 103 = 20 V
Electrical power and energy
Hence,
Electrical power
power P = V × I = 20 × 4 × 10−3
Power P in an electrical circuit is given by the product of potential difference V and current I, as stated in
Chapter 1. The unit of power is the watt, W.
P = V × I watts
Hence
(1)
From Ohm’s law, V = IR. Substituting for V in equation (1) gives:
= 80 mW
Problem 11. An electric kettle has a resistance of
30 . What current will flow when it is connected
to a 240 V supply? Find also the power rating of the
kettle.
P = (IR) × I
Current, I =
P = I 2 R watts
i.e.
Power, P = VI = 240 × 8 = 1920 W
Also, from Ohm’s law, I = V /R. Substituting for I in
equation (1) gives:
V
P=V×
R
V2
i.e.
P=
watts
R
There are thus three possible formulae which may be
used for calculating power.
= 1.92 kW = power rating of kettle
Problem 12. A current of 5 A flows in the winding
of an electric motor, the resistance of the winding
being 100 . Determine (a) the p.d. across the
winding, and (b) the power dissipated by the coil.
(a)
Problem 9. A 100 W electric light bulb is
connected to a 250 V supply. Determine (a) the
current flowing in the bulb, and (b) the resistance of
the bulb.
Power P = V × I, from which, current I =
(a)
100 10 2
Current I =
=
= = 0.4 A
250 25 5
(b)
Resistance R =
Power P = I R = (4 × 10
3
= 16 × 10−6 × 5 × 103
= 0.08 W or 80 mW
(b)
Power dissipated by coil,
P = I 2 R = 52 × 100
= 2500 W or 2.5 kW)
) (5 × 10 )
= 80 × 10−3
V = IR = 5 × 100 = 500 V
(Alternatively, P = V × I = 500 × 5
V 250 2500
=
=
= 625 I
0.4
4
−3 2
Potential difference across winding,
= 2500 W or 2.5 kW
P
V
Problem 10. Calculate the power dissipated when
a current of 4 mA flows through a resistance of
5 k.
2
V
240
=
= 8A
R
30
Problem 13. The hot resistance of a 240 V
filament lamp is 960 . Find the current taken by
the lamp and its power rating.
From Ohm’s law,
V
240
=
R
960
1
24
= A or 0.25 A
=
96
4
Power rating P = VI = (240) 41 = 60 W
current I =
Section 1
An introduction to electric circuits 15